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∫sinx^2Dx

定积分∫sinx^2dx=∫0.5*(1-cos2x)dx=x-0.25sin2x+C

是0,因为后面的求导对象是一个定积分,对于求导的变量x来说是一个常数(积分值取决于被积分函数和积分限),相当于对常数求导,至于后面积分变量是x其实和前面求导的x没有关系,换成任何字母都一样,其值取决于积分限a、b(被积分函数固定了,...

∫x(sinx)^2dx =(1/2)∫x(1-cos2x)dx =(1/4)x^2-(1/2)∫xcos2xdx =(1/4)x^2-(1/4)∫xdsin2x =(1/4)x^2-(1/4)xsin2x +(1/4)∫sin2x dx =(1/4)x^2-(1/4)xsin2x -(1/8)cos2x + C

因为(xsinx)2=x2(1?cos2x)2,所以∫π0(xsinx)2dx =∫π0x22dx-∫π0x2cos2x2dx.利用分部积分法可得,∫π0x2cos2x2dx=x2sin2x4|π0-∫π0xsin2x2dx=0-(?xcos2x4|π0+∫π0cos2x4dx)=-π4-sin2x8|π0=-π4,又因为 ∫π0x22dx=x36|π0=π36,所以∫π0(xsinx)2dx=

令arcsin(x/2)=u,则x=2sinu ∫arcsin(x/2)dx =∫ud(2sinu) =2∫ud(sinu) =2usinu-2∫sinudu =2usinu+2cosu +C =x·arcsin(x/2) +√(4-x²) +C

若是 ∫ xsin(x^2)dx 则 ∫ xsin(x^2)dx = (1/2)∫ sin(x^2)d(x^2) = - (1/2)[cosx^2] = (1/2)[1-cos(π^2/4)] 若是 ∫ x(sinx)^2dx 则 ∫ x(sinx)^2dx = (1/2) ∫ x(1-cos2x)dx = (1/2) ∫ xdx - (1/2) ∫ xcos2xdx = (1/4) [x^2] - (1/4) ∫ xdsin2x = π...

x^5sinx^2dx =x^2*sinx^2-1/2*x^4*cosx^2+cosx^2 如果你认可我的回答,敬请及时采纳, ~如果你认可我的回答,请及时点击【采纳为满意回答】按钮 ~~手机提问的朋友在客户端右上角评价点【满意】即可。 ~你的采纳是我前进的动力 ~~O(∩_∩)O,记得好...

解: 令arcsinx=u,则x=sinu ∫(arcsinu/u²)du =∫(u/sin²u)d(sinu) =-∫[ud(1/sinu) =-u/sinu +∫cscudu =-u/sinu +ln|cscu-cotu| +C =-(arcsinx)/x +ln|[1-√(1-x²)]/x| +C

如下

方法一: ∫[1/(sinx+cosx)^2]dx =(1/2)∫{1/[(1/√2)sinx+(1/√2)cosx]^2}dx =(1/2)∫{1/[sinxcos(π/4)+cosxsin(π/4)]^2}dx =(1/2)∫{1/[sin(x+π/4)]^2}d(x+π/4) =-(1/2)cot(x+π/4)+C 方法二: ∵1...

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